Question

Evaluate the integral? : `int_0^(pi/6) sinx/cos^2x dx`

Answer 1

`int_0^(pi/6) \ sinx/cos^2x \ dx = 2/sqrt(3) - 1`

Explanation:

We seek:

`I = int_0^(pi/6) \ sinx/cos^2x \ dx`

We can perform a simple substitution:

Let `u=cosx => (du)/dx = -sinx`

And we must change the limits of integration:

When `x= { (0), (pi/6) :} => u = { (1), (sqrt(3)/2) :}`

So substituting into the integral, we get:

`I = int_0^(pi/6) \ (-1)/cos^2x \ (-sinx) \ dx`
`\ \ = - int_1^(sqrt(3)/2) \ 1/u^2 \ du`
`\ \ = [1/u]_1^(sqrt(3)/2)`
`\ \ = 1/(sqrt(3)/2) - 1/1`
`\ \ = 2/sqrt(3) - 1`